package sword2offer;

/**
 * <p>
 * 合并两个有序链表
 * </p>
 * 将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
 * @author LovelyBigDuck
 * @date 2022/3/26 23:29
 */
public class S_21 {

    public class ListNode {
        int val;
        ListNode next;
        ListNode() {}
        ListNode(int val) { this.val = val; }
        ListNode(int val, ListNode next) { this.val = val; this.next = next; }
    }

    // 递归
    class Solution {
        public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
            if (l1 == null) {
                return l2;
            } else if (l2 == null) {
                return l1;
            } else if (l1.val < l2.val) {
                l1.next = mergeTwoLists(l1.next, l2);
                return l1;
            } else {
                l2.next = mergeTwoLists(l1, l2.next);
                return l2;
            }
        }
    }

    // 迭代
    class Solution1 {
        public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
            ListNode prehead = new ListNode(-1);

            ListNode prev = prehead;
            while (l1 != null && l2 != null) {
                if (l1.val <= l2.val) {
                    prev.next = l1;
                    l1 = l1.next;
                } else {
                    prev.next = l2;
                    l2 = l2.next;
                }
                prev = prev.next;
            }

            // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
            prev.next = l1 == null ? l2 : l1;

            return prehead.next;
        }
    }
}
